Question

What is the heat required to boil 743 grams of water in a kettle at 100c

Answer 1

Solution:Given:

$m=743g$

$c=2.02\;(\frac{J}{g\;C})$

$\Delta T=100^oC$

$Q=mc\Delta T=(743g)(2.02 \;(\frac{J}{g\;C}))(100^oC)=150086J=150.086kJ$